$f:X\to Y$, $A,B,\subseteq X$. Show that $f(A\setminus B)=f(A)\setminus f(B)$ iff $f(A\setminus B)\cap f(B) =\emptyset$ I tried to prove this but I am not sure if its correct. Please help me out with any tips or advice on how to improve. Here it is:
First let $f(A\setminus B)\cap f(B)=\emptyset$. Now $$f(A\setminus B)=(f(A)\setminus f(B)) \cup \{     y\in Y| \exists (x_1 \in (A\setminus B), x_2\in B) |f(x_1)=f(x_2 ) \}.$$
Now by the hypothesis $f(A\setminus B)\cap f(B)=\emptyset$ we have that $\nexists (x_1 \in (A\setminus B), x_2\in B) |f(x_1)=f(x_2 ) $. Thus
$$f(A\setminus B)=(f(A)\setminus f(B)) \cup \emptyset=f(A)\setminus f(B).$$
Now for the reverse implication we assume that $f(A\setminus B)=f(A)\setminus f(B)$. We intersect both sides with $f(B)$ to obtain
$$f(A\setminus B)\cap f(B)=(f(A)\setminus f(B))\cap f(B)=\emptyset.$$
So this is my proof. I personally think it's correct but very unclear possible. Any help or verification will be appreciated. Thanks in advance
 A: Here is another full proof, relying more on the symbols than the earlier answers, but therefore more mechanical as well: start at the most complex side, expand the definitions, simplify, and take it from there.
We calculate as follows:
\begin{align}
& f[A \setminus B] = f[A] \setminus f[B] \\
= & \qquad \text{"set extensionalify; definition of $\;\setminus\;$"} \\
& \langle \forall y :: y \in f[A \setminus B] \;\equiv\; y \in f[A] \land \lnot (y \in f[B]) \rangle \\
= & \qquad \text{"basic property of $\;\cdot[\cdot]\;$, three times"} \\
& \langle \forall y :: \langle \exists x : f(x) = y : x \in A \land \lnot (x \in B) \rangle
  \\&\phantom{\langle \forall y :: } \equiv\;
  \langle \exists x : f(x) = y : x \in A \rangle
  \land \lnot \langle \exists x : f(x) = y : x \in B \rangle \rangle \\
= & \qquad \text{"the last part is equivalent to $\;\langle \forall x : f(x) = y : \lnot(x \in B) \rangle\;$ by} \\
& \qquad \phantom{\text{"}}\text{DeMorgan: use this to add $\;\lnot(x \in B)\;$ on other side of $\;\land\;$ -- this} \\
& \qquad \phantom{\text{"}}\text{makes the right hand side of $\;\equiv\;$ more similar to the left hand side"} \\
& \langle \forall y :: \langle \exists x : f(x) = y : x \in A \land \lnot (x \in B) \rangle
  \\&\phantom{\langle \forall y :: } \equiv\;
  \langle \exists x : f(x) = y : x \in A \land \lnot (x \in B) \rangle
  \land \lnot \langle \exists x : f(x) = y : x \in B \rangle \rangle \\
= & \qquad \text{"logic: simplify $\;P \equiv P \land \lnot Q\;$ to $\;\lnot (P \land Q)\;$"} \\
& \langle \forall y :: \lnot (
  \langle \exists x : f(x) = y : x \in A \land \lnot (x \in B) \rangle
  \land \langle \exists x : f(x) = y : x \in B \rangle
  ) \rangle \\
= & \qquad \text{"basic property of $\;\cdot[\cdot]\;$, twice"} \\
& \langle \forall y :: \lnot (y \in f[A \setminus B] \land y \in f[B]) \rangle \\
= & \qquad \text{"definition of $\;\cap\;$; basic property of $\;\emptyset\;$"} \\
& f[A \setminus B] \cap f[B] = \emptyset \\
\end{align}
This completes the proof.
If you want to see the above logical simplification in more detail, we have
\begin{align}
& P \;\equiv\; P \land \lnot Q \\
= & \qquad \text{"$\;p \equiv p \land q\;$ is one of the ways to write $\;p \Rightarrow q\;$"} \\
& P \;\Rightarrow\; \lnot Q \\
= & \qquad \text{"$\;\lnot p \lor q\;$ is another way to write $\;p \Rightarrow q\;$"} \\
& \lnot P \lor \lnot Q \\
= & \qquad \text{"DeMorgan"} \\
& \lnot (P \land Q) \\
\end{align}
A: Assume that $f\left(A\backslash B\right)=f\left(A\right)\backslash f\left(B\right)$.
If $y\in f\left(A\backslash B\right)$ then $y\in f\left(A\right)\backslash f\left(B\right)$
so that $y\notin f\left(B\right)$ this proves that $f\left(A\backslash B\right)\cap f\left(B\right)=\emptyset$.
Assume that $f\left(A\backslash B\right)\cap f\left(B\right)=\emptyset$
or equivalently $f\left(A\backslash B\right)\subset f\left(B\right)^{c}$.
From $A\backslash B\subset A$ it follows that $f\left(A\backslash B\right)\subset f\left(A\right)$.
Then $f\left(A\backslash B\right)\subset f\left(A\right)\cap f\left(B\right)^{c}=f\left(A\right)\backslash f\left(B\right)$.
Conversely, if $y\in f\left(A\right)\backslash f\left(B\right)$ then
$y=f\left(x\right)$ for some $x\in A$, and $x\in B$ cannot be true (because that would imply that $y\in f\left(B\right)$). So
$x\in A/B$ and consequently $y=f\left(x\right)\in f\left(A\backslash B\right)$. This proves that $f\left(A\right)\backslash f\left(B\right)\subset f\left(A\backslash B\right)$.
A: The first part of the proof is very unclear to me. I personally would like to show that both sets are subsets of each other to prove a set equality. Here's my humble suggestion:
Suppose $f(A \setminus B)∩f(B)= \emptyset$:
$y \in f(A \setminus B) \implies \exists  x \in (A \setminus B)$ such that $f(x) = y$. Since $x \in A$, $ y = f(x) \in f(A)$. Furthermore due to our hypothesis, $y \in f(A \setminus B) \implies y \notin f(B) $. 
Since $y \in f(A)$ and $y \notin f(B)$, $y \in f(A) \setminus f(B)$$\implies f(A \setminus B) \subseteq f(A) \setminus f(B)$ --------- $(1)$ 
Now say $y \in f(A) \setminus f(B)$. This says $\exists x \in A$ such that $y = f(x)$ and there is no $x' \in B$ such that $y = f(x')$. Clearly $x \in A$ and $x \notin B \implies x \in A \setminus B \implies y = f(x) \in f(A \setminus B)$
This implies $f(A) \setminus f(B) \subseteq f(A \setminus B) $ ----------------- $(2)$
Equations $(1)$ and $(2)$ provide the set equality. 
The second part of the proof is impressive by the way..
A: Hint:


*

*Use $$X \setminus Y = X \cap Y^c,$$ and $$\big(f(B)\big)^c \cap f(B \cup B^c) \subseteq f(B^c). \tag{$\spadesuit$}$$

*$(\Rightarrow)$ From the assumption we get $f(A \cap B^c) = f(A) \cap \big(f(B)\big)^c \subseteq \big(f(B)\big)^c.$

*$(\Leftarrow)$ From $f(A \cap B^c) \subseteq \big(f(B)\big)^c$ we get $\subseteq$ and from $(\spadesuit)$ we get $\supseteq$ .


I hope this helps $\ddot\smile$
